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Fiesta28 [93]
2 years ago
12

The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years. Based on th

e Central Limit Theorem we know the distribution of means from every sample of size 60 will be , with a mean of and a standard deviation of . The probability that a sample mean is 12 or larger for a sample from the horse population is_________
Mathematics
1 answer:
elena-14-01-66 [18.8K]2 years ago
3 0

Answer: 0.9013

Step-by-step explanation:

Given mean, u = 10, standard deviation =8

P(X) =P(Z= X - u /S)

We are to find P(X> or =12)

P(X> or = 12) = P(Z> 12-10/8)

P(Z>=2/8) = P(Z >=0.25)

P(Z) = 1 - P(Z<= 0.25)

We read off Z= 0.25 from the normal distribution table

P(Z) = 1 - 0.0987 = 0.9013

Therefore P(X> or=12) = 0.9013

Note the question was given as an incomplete question the correct and complete question had to be searched online via Google. So the data used are those gotten from the online the Googled question.

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Which statements about the figure are true? Select two options.
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<u>Complete Question</u>

The circle is inscribed in triangle PRT. A circle is inscribed in triangle P R T. Points Q, S, and U of the circle are on the sides of the triangle. Point Q is on side P R, point S is on side R T, and point U is on side P T. The length of R S is 5, the length of P U is 8, and the length of U T is 6. Which statements about the figure are true?

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The diagram of the question is drawn for more understanding,

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The owner of a manufacturing plant employs eighty people. As part of their personnel file, she asked each one to record to the n
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Answer:

<em>Mean of the sample = 27.83</em>

<em> The variance of the the sample = 106.96</em>

<em> </em><em>Standard deviation of the sample = 10.34</em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given random sample of six employees

x     26    32    29    16     45    19

mean of the sample

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Mean of the given data = 27.83

<u>Step(ii):-</u>

<u>Given data</u>

x           :       26         32           29           16           45        19

x - x⁻     :      -1.83      4.17        1.17       -11.83     17.17    -8.83

(x - x⁻)²  :    3.3489   17.3889   1.3689   139.9489  294.80 77.9689

∑ (x-x⁻)²  =   534.8245

Given sample size 'n' =6

The variance of given data

           S²  = ∑(x-x⁻)² / n-1  

           S^{2}  = \frac{534.8245}{6-1} = 106.9649

The variance of the given sample = 106.9649

<u> Step(iii):-</u>

Standard deviation of the given data

S = \sqrt{variance} = \sqrt{106.9649} =10.3423

Standard deviation of the sample = 10.3423

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